Question: Let $f(x) = -x^{2}+10x+4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Answer: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-x^{2}+10x+4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -1, b = 10, c = 4$ $ x = \dfrac{-10 \pm \sqrt{10^{2} - 4 \cdot -1 \cdot 4}}{2 \cdot -1}$ $ x = \dfrac{-10 \pm \sqrt{116}}{-2}$ $ x = \dfrac{-10 \pm 2\sqrt{29}}{-2}$ $x =5 \pm \sqrt{29}$